ARNOLD’S ELEMENTARY PROOF OF THE INSOLVABILITY OF THE QUINTIC
LEO GOLDMAKHER
A BSTRACT. We give a proof (due to Arnold) that there is no quintic formula. Somewhat more precisely, we show
that any finite combination of the four field operations (+, −, ×, ÷), radicals, the trigonometric functions, and the
exponential function will never produce a formula for producing a root of a general quintic polynomial. The proof
is elementary, requiring no knowledge of abstract group theory or Galois theory.
1. P REREQUISITE IDEAS AND NOTATIONS
To understand the arguments in this essay you don’t need to know Galois theory. You also don’t need to know
abstract algebra or group theory. However, you do need to know about complex numbers and the complex plane
C. Let’s begin with a basic
Question. What is the definition of the imaginary number i?
We all know the answer: it’s the square-root of −1. There’s only one problem. There are two square roots of
−1! Thus, our definition defines two numbers ±i but doesn’t give a way to distinguish between them. So how
can we define i without simultaneously defining −i?
It turns out that this is impossible to do.1 Of course, we can arbitrarily label one of the two square-roots as i,
and then this forces the other to be −i. But this choice is arbitrary, and demonstrates that there’s a fundamental
symmetry between the two numbers ±i; the two are yoked together and are algebraically indistinguishable.
This observation lies at the heart of Galois theory, which studies other symmetries between numbers (not just
pairs of numbers, but also triples and quadruples etc.) and, using group theory to extract properties of the
symmetries, deduces properties that such numbers must have.2
These types of ideas will play a major role in our discussion below, as well. Unlike Galois theory, we won’t
require abstract group theory, but we will use a convenient notation – cycle notation – to describe permutations
of objects. Rather than defining this formally, let’s consider an example. Suppose you have six objects and you
want to rearrange them so that
•
•
•
•
•
•
the first object ends up in the third position,
the second object ends up in the sixth position,
the third object ends up in the fourth position,
the fourth object ends up in the first position,
the fifth object doesn’t move, and
the sixth object ends up in the second position.
In cycle notation, this permutation is written (1 3 4)(2 6). Note that 5 doesn’t appear here at all, since it ends
up being fixed. The trivial permutation, denoted (), is the one which leaves all the objects where they are.
One nice feature of cycle notation is that it’s very easy to undo a given permutation: just write the numbers in
reverse order inside each set of parentheses. For example, the permutation which undoes the one given above
is (4 3 1)(6 2). Note that (4 3 1) = (1 4 3) and (6 2) = (2 6), so we could have also written the overall
permutation in the form (1 4 3)(2 6). To learn more about cycle notation, check out the relevant subsection in
Wikipedia’s article on permutations.
Enough preliminaries. Let’s do some math!
1
Here I’m restricting the term define to mean solving polynomials whose coefficients are real numbers.
More precisely, in Galois theory one derives information about how numbers can (and can’t) be described.
2
2. Q UADRATIC FORMULAS AREN ’ T FUNCTIONS
The first step to understanding Arnold’s idea is to build up some intuition. Navigate to the website
duetosymmetry.com/tool/polynomial-roots-toy/
Set the degree of the polynomial to 2. There are two copies of C drawn. On the left is coefficient space –
there are two dots, labelled a0 and a1 , corresponding to the coefficients of a quadratic polynomial (precisely,
the polynomial x2 + a1 x + a0 ). The C on the right is root space: the two dots displayed are the two roots of
the polynomial defined by the current state coefficient space.
To get a feel for all this, drag the a0 coefficient to −1 and the a1 coefficient to 1/2. You should have two real
roots in root space (one at ≈ −1.28, the other at ≈ 0.78). Let’s call r1 the negative root, and r2 the positive
root. Now move the coefficient a0 around in a small loop (i.e. move it around a little bit, and then return it to
−1 where it started). Note that the roots move continuously, and then return to their original positions. Next,
move a0 in a big loop (big enough that it orbits around r2 ). Something funny happens: the roots r1 and r2
switch places.
Pause and think about this for a second. This is really, really weird. Here’s one immediate consequence of
this observation:
Proposition 1. There does not exist any continuous function from the space of quadratic polynomials to C
which associates to any quadratic polynomial a root of that polynomial.
Proof. Exercise.
Wait, what? Don’t we have a quadratic formula?! Well, here’s the thing: the quadratic formula is
√
−b ± b2 − 4ac
2a
which isn’t a function, since it outputs two values for a given input. You might√think I’m cheating by writing
the ± in there. What if we just write +? Well, it’s still not a function, because z isn’t a function. We saw an
example of this at the very beginning of this essay, when discussing the definition of i. But even
√ in cases when
we’re used to being able to canonically choose the value of√
the squareroot – for example, 1 – there’s more
than meets the eye. To convince yourself of this, consider e2πit . No matter how we choose√to define this,
2πit := eπit .
we end up with a problem
√ if we allow t to vary broadly enough. For example,
√ suppose we set e
Plugging in t = 0 gives
1 = 1, unsurprisingly. But plugging in t = 1 gives 1 = −1.
√
Thus, we see that z isn’t a function over C. By contrast, it turns out that sin, cos, and exp are all continuous
functions over C. In particular, Proposition 1 implies:
Corollary 2. There’s no quadratic formula built out of a finite combination of +, −, ×, ÷, the functions
sin, cos, exp, and the coefficients of the polynomial.
3. C ONNECTING LOOPS AND PERMUTATIONS
Let’s go back to the website and play more with coefficients and roots. Set the degree of the polynomial to
be three, and maneuver the coefficients so that
1 1
1 1
a0 = − + i
a1 = 0
and
a2 = − − i.
2 2
2 2
This produces three roots, which we label as
1 1
r1 = − − i
r2 = i
and
r3 = 1.
2 2
Moving a0 around in different loops (making sure it ends up where it started), we see that we can induce at
least two different permutations on the roots: the trivial permutation () which fixes all three, and the nontrivial
permutation (1 2 3) which moves r1 to second position, r2 to third position, and r3 to first position. Can we induce other permutations by moving around the other coefficients? For example, can we induce the permutation
(1 2)?
Sure. The trick is to move the roots around, and track what happens to the coefficients. Let’s manually induce
the permutation (1 2) by creating two (non-intersecting) paths connecting r1 and r2 in root space. Then we
simultaneously move r1 along one of the paths, and r2 along the other, until they switch places. As we do this,
we keep track of how the coefficients move. The key point is that the coefficients move in a continuous way,
and they all return to their initial positions. They have to, since at the end of the day we have the same three
roots as we started with (just permutated). Thus, there exist three loops in coefficient space – each based at one
of the ai ’s – which, when we move the coefficients around them simultaneously, induce the permutation (1 2)
on the roots. Similarly, we can induce any permutation of the roots we wish.
We can simplify the discussion by viewing the three coefficients as the coordinates of a single point. In other
words, rather than coefficient space, we form function space C3 , where each point (b, c, d) corresponds to a
unique cubic x3 + bx2 + cx + d. This allows us to rephrase the conclusion of the previous paragraph as follows.
Proposition 3. Pick any three distinct points r1 , r2 , r3 ∈ C, and let p denote the point of C3 corresponding to
the polynomial (x − r1 )(x − r2 )(x − r3 ). For any permutation of the ri there exists a loop in C3 based at p
which induces this permutation.
Note that even though C3 contains all cubic polynomials (up to a constant factor), the polynomial p in this
proposition is of a special kind: all its roots are distinct. We’ll be dealing with this type of polynomial a lot, so
we give it a name:
Definition. A polynomial of degree n is called separable if and only if it has n distinct roots.
This notion allows us to restate Proposition 3 so that we are given the polynomial (as opposed to its roots). Let
F3 denote the set of all points in function space C3 which correspond to separable cubic polynomials.
Proposition 4. Pick any p ∈ F3 and denote the roots of the corresponding polynomial by r1 , r2 , r3 . For any
permutation of the ri there exists a loop in F3 based at p which induces this permutation.
More generally, set
Fn := {(a1 , a2 , . . . , an ) ∈ Cn : xn + a1 xn−1 + a2 xn−2 + · · · + a0 is separable};
call this n-dimensional function space. Then we have
Proposition 5. Pick any p ∈ Fn and denote the roots of the corresponding polynomial by r1 , r2 , . . . , rn . For
any permutation of the ri there exists a loop in Fn based at p which induces this permutation.
This is one of the three building blocks of our proof of the insolvability of the quintic. Follow me, dear reader!
4. C UBIC FORMULAS REQUIRE NESTED RADICALS
Let’s return to F3 (the space of separable cubic polynomials). Pick any p ∈ F3 , and denote its roots r1 , r2 , r3 .
By Proposition 5, we can find a loop γ in function space which induces the permutation (1 2) in root space.
Since this is a nontrivial permutation, we arrive at the same conclusion as Proposition 1: there does not exist any
continuous function which produces a root of every cubic polynomial. Given that this held even for quadratic
polynomials, this is hardly surprising. The goal of this section is to go further: we’ll show that even if we allow
radicals to be used (in combination with continuous functions), any cubic
√ formula requires nested radicals.
· is not a loop:p
the loop γp
: [0, 1] → C
We’ve already seen an example of a loop in C whose image under
p
defined by γ(t) := e2πit . No matter how we choose to define γ(0), we find that γ(0)
=
6
γ(1), even
√
though γ(0) = 1 = γ(1). However, there are some types of loops whose image under · is a loop. Here’s
a trivial way to construct such a loop: given any loop γ based at p, denote by γ −1 the loop based at p which
follows the same path as γ but in reverse. It’s not too hard to √
prove that the image of the composite loop γγ −1
−1
(by which I mean: go around γ, then go around γ ) under · is a loop. A more interesting version of such
back and forth travel is called the commutator:
Definition. The commutator of two loops γ1 , γ2 (both based at the same point) is defined to be
[γ1 , γ2 ] := γ1 γ2 γ1−1 γ2−1 .
In words, the commutator means: go around γ1 , then around γ2 , then backwards around γ1 , then backwards
along γ2 .
The reason we care about commutator loops is that they’re trivial enough that their image under radicals is
nice, but complicated enough that they can induce nontrivial permutations of roots. We expand on these two
points. First:
Exercise 1. Suppose γ1 and γ2 are two loops based at the same point in Fn , and pick any continuous function
f : Fn → C. Then for any α ∈ Q the image of f (p)α as p traverses the loop [γ1 , γ2 ] is a loop in C.
Next, we show that commutator loops can induce nontrivial permutations of roots. More precisely, pick some
separable cubic polynomial p. By Proposition 5 there exists some loop γ1 in function space F3 which induces
the permutation (1 2 3) on the roots of p, and some other loop γ2 in F3 which induces the permutation (1 2) on
the roots of p.
Exercise 2. What permutation does the commutator loop [γ1 , γ2 ] induce on the roots of p? In particular, show
that no root is left fixed.
This immediately implies that the image of [γ1 , γ2 ] under any cubic formula cannot be a loop in C! By contrast,
Exercise 1 implies that any combination of the four field operations, continuous functions, and a single nesting
of radicals would produce a loop. We have therefore proved
Proposition 6. Any cubic formula built solely out of field operations, continuous functions, and radicals must
contain nested radicals.
5. I NSOLVABILITY OF THE QUINTIC
In this section we will adapt our arguments from the previous section to prove
Corollary 7. There does not exist any quintic formula built out of a finite combination of field operations,
continuous functions, and radicals.
The inclusion of the word finite above is very important. For example:
Exercise 3. Express a solution to x5 − x − 1 = 0 using just +, ×, and infinitely many nested radicals.
Our proof of Corollary 7 will look quite similar to that of Proposition 6. We begin with
Proposition 8. There exists a commutator loop which induces the permutation (1 2 3 4 5).
There are many ways to prove this, the most direct of which is trial-and-error. Although not the most elegant
solution, this makes explicit that we don’t need any knowledge of abstract group theory for the proof. Here’s
the approach.
Given any two permutations of five elements, and any separable quintic polynomial p, Proposition 5 guarantees the existence of two loops γ1 and γ2 in function space F5 (based at the polynomial p) which induce
these two permutations on the roots of p. Now, the commutator loop [γ1 , γ2 ] induces the commutator of the two
corresponding permutations. Thus, given any permutation which is a commutator of two permutations, there
exists a commutator loop inducing this permutation. We are thus led to
Question. Which permutations on five elements can be expressed as the commutator of two permutations on
five elements?
There are 120 different permutations of five elements. If we compute all 14400 = (120)2 commutators of pairs
of permutations (not so hard to program a computer to do), we see that
• there are precisely 60 different permutations which can be formed as commutators of permutations, and
• (1 2 3 4 5) is one of these.
As was the case with cubics, right away this tells us that any quintic formula built out of the field operations,
continuous functions, and radicals must have nested radicals.
But now we go a step further. Consider the 60 different commutator permutations, and look at the 3600
possible commutators of any pair of these. It turns out we get the same 60 permutations we started with! In
particular, (1 2 3 4 5) is a commutator of commutators, which means that any quintic formula built out of the
field operations, continuous functions, and radicals must have nested, nested radicals. Clearly, we need some
better way to discuss this. Let’s say that an expression built out of the field operations, continuous functions,
and radicals has nesting of level k iff the expression is of the form
s
r
q
p
a1
a2
a3
f1 +
f2 +
· · · + ak fk .
In this language, we see that any quintic formula built out of the field operations, continuous functions, and
radicals must have nesting of level ≥ 3.
Since we can keep taking commutators of commutators of commutators of... and still have the same 60 left,
including (1 2 3 4 5), we can iterate the argument about to conclude that
Theorem 9 (Arnold, 1963?). Fix any positive integer N . Any quintic formula built out of the field operations,
continuous functions, and radicals must have nesting of level ≥ N .
Corollary 7 immediately follows.
One important point is that we are not disproving the possibility of solving any particular quintic using finitely
many symbols. For example, we can easily express a solution to the quintic equation
x5 − 1 = 0.
Instead, we proved the impossibility of a formula which produces a root for arbitrary separable quintics.
Exercise 4. Where in the proof did we require the quintic formula to be general? In other words, why doesn’t
the proof apply to a particular quintic?
6. H ISTORY AND RELATION TO G ALOIS THEORY
The ultimate goal of a typical introductory course on Galois theory is to develop an algorithm which, given a
polynomial, outputs a certain group (the ‘Galois group of the polynomial’). The shape of this group then tells
you about the shape of the roots of the polynomial, in particular indicating the level of nesting of radicals. For
example, the Galois algorithm proves that any root of x5 − x − 1 requires infinite nesting of radicals to write
down. In this sense, Galois theory says more than Arnold’s theorem (our Theorem 9), which only asserts that
any general quintic formula must have infinite nesting. However, in another sense, Arnold’s approach gives a
stronger result than Galois theory: Theorem 9 allows not only the use of radicals but also of sin(), exp(), and
any other continuous function.
The other advantage of Arnold’s approach is immediately evident to anyone who’s taken a course on Galois
theory: it’s much more straightforward. In Galois theory, the algorithm comes out of a semester’s worth
of difficult proofs... not to mention that figuring out the Galois group of a given polynomial can be highly
nontrivial!
Although Arnold’s proof is surprisingly elementary, it should not come as a shock that one can prove such
a theorem without Galois theory, since the insolvability of the quintic was originally proved before Galois. In
1799 – about 250 years after the discovery of the quartic formula – Paolo Ruffini announced a proof that no
general quintic formula exists. However, his proof was 500 pages long (and, as it turned out later, had logical
gaps). In 1824, the 22-year-old Norwegian genius Niels Henrik Abel published a six-page proof of Ruffini’s
assertion; an expanded version appeared two years later in Crelle’s journal.3 Neither Abel’s nor Ruffini’s proofs
relied on Galois groups, since Galois’ paper on the subject was only submitted in 1830, and not published until
1846. A wonderful account of Abel’s proof and the history can be found in Rosen’s essay [Ros95].
I wrote this essay because I was unable to find a suitable written version of Arnold’s proof. The most famous
written version is the problem book [Ale04], and a more standard description is given in the paper [Zol00].
3
Apparently Gauss ignored an early manuscript Abel sent him, dismissing it as the work of a crank.
However, the former is hundreds of pages long, while the latter is short but written for a mathematically sophisticated reader. The goal of the present manuscript is meant to preserve the spirit of Arnold’s original goal
in devising the proof: to make this beautiful theorem accessible to a non-professional mathematician.4
Acknowledgements. I am indebted to Boaz Katz, whose beautiful video explanation
https://www.youtube.com/watch?v=RhpVSV6iCko
exposed me to Arnold’s proof and inspired me to write this document. I’m also grateful to Leo C. Stein for
creating and sharing his tool for exploring the interaction between coefficients and roots:
duetosymmetry.com/tool/polynomial-roots-toy/
R EFERENCES
[Ale04] V. B. Alekseev, Abel’s theorem in problems and solutions, Kluwer Academic Publishers, Dordrecht, 2004. Based on the
lectures of Professor V. I. Arnold; With a preface and an appendix by Arnold and an appendix by A. Khovanskii.
[Ros95] Michael I. Rosen, Niels Hendrik Abel and equations of the fifth degree, Amer. Math. Monthly 102 (1995), no. 6, 495–505.
[Zol00] Henryk Zoladek, The topological proof of Abel-Ruffini theorem, Topol. Methods Nonlinear Anal. 16 (2000), no. 2, 253–265.
D EPT OF M ATHEMATICS & S TATISTICS , W ILLIAMS C OLLEGE , W ILLIAMSTOWN , MA, USA
E-mail address: leo.goldmakher@williams.edu
4
Arnold’s goal was to make it accessible to high school students, while this document is probably better-suited to undergraduate
math majors.